Rolle's Theorem Questions and Examples

Rolle's theorem is a special case of the mean value theorem. It is discussed here through examples and questions.

Rolle's Theorem

Rolle's theorem is the result of the mean value theorem where under the conditions:
\( f(x) \) be a continuous functions on the interval \( [a, b] \) and differentiable on the open interval \( (a, b) \) , there exists at least one value c of x such that \[ f '(c) = \dfrac{f(b) - f(a)}{b - a} \] Now if the condition \( f(a) = f(b) \) is satisfied, then the above simplifies to : \( f'(c) = 0 \).
In other words, under the above three conditions we can always find a tangent to the curve of f that is horizontal (slope = \( f'(c) = 0 \) ).
Rolle's Theorem

If \( f(x) \) is
1) a continuous function on the interval \( [a, b] \)
2) differentiable on the open interval \( (a, b) \)
3) and \( f(a) = f(b) \),
then there is at least one value c of x in the interval \( (a, b) \) such that
\( f'(c) = 0 \)

Example 1

The graph of \( f(x) = - x^2 + 6x - 6 \) for \( 1 \leq x \leq 5 \) is shown below. \( f(1) = f(5) = - 1 \) and \( f \) is continuous on [1 , 5] and differentiable on (1 , 5) hence, according to Rolle's theorem, there exists at least one value of \( x = c \) such that \( f '(c) = 0 \).
\( f '(x) = - 2 x + 6 \)
\( f '(c) = - 2 c + 6 = 0 \)
Solve the above equation to obtain
\( c = 3 \)
Therefore at \( x = 3 \) there is a tangent to the graph of \( f \) that has a slope equal to zero (horizontal line) as shown in figure 1 below.

Example 2

The graph of \( \quad f(x) = \sin(x) + 2 \) for \( \quad 0 \leq x \leq 2\pi \) is shown below. \( \quad f(0) = f(2\pi) = 2 \) and \( f \) is continuous on [0 , \( 2\pi \)] and differentiable on (0 , \( 2\pi \)) hence, according to Rolle's theorem, there exists at least one value (there may be more than one!) of \( x = c \) such that \( f '(c) = 0 \).
\( f '(x) = \cos(x) \)
\( f '(c) = \cos(c) = 0 \)
The above equation has two solutions on the interval [0 , \( 2\pi \)]
\( c_1 = \dfrac{\pi}{2} \) and \( c_2 = \dfrac{3\pi}{2} \).
Therefore both at \( x = \dfrac{\pi}{2} \) and \( x = \dfrac{3\pi}{2} \) there are tangents to the graph that have a slope equal to zero (horizontal line) as shown in figure 2 below.

Example 3

Function \( f \) in figure 3 does not satisfy Rolle's theorem: although it is continuous and \( f(-1) = f(3) \), the function is not differentiable at \( x = 1 \) and therefore \( f '(c) = 0 \) with \( c \) in the interval (-1 , 3) is not guaranteed. In fact it is easy to see that there is no horizontal tangent to the graph of \( f \) on the interval (-1 , 3).

Graphs of a function with at least one of the conditions not satisfied — Figure 3. Graph of function that does not satisfy the condition of differentiability in Rolle's theorem

Question 1

Which of the functions given below satisfy all three conditions of Rolle's theorem?
a) \( f(x) = \cos(x) \), for \( x \) in \( [0 , 2\pi \)] \)
b) \( g(x) = |x - 2| \), for \( x \) in \( [0 , 4] \)
c) \( h(x) = 1 / x^2 \), for \( x \) in \( [-1 , 1] \)
d) \( k(x) = |\sin(x)| \), for \( x \) in \( [0 , 2\pi ] \)

solution to question 1

a)
\( f(0) = 1 \) and \( f(2\pi) = 1 \) therefore \( f(0) = f(2\pi) \)
\( f \) is continuous on \( [0 , 2\pi ] \)
Function \( f \) is differentiable in \( (0 , 2\pi ) \)
Function \( f \) satisfies all conditions of Rolle's theorem
b)
function \( g \) has a V-shaped graph with vertex at \( x = 2 \) and is therefore not differentiable at \( x = 2 \).
Function \( g \) does not satisfy all conditions of Rolle's theorem
c)
Function \( h \) is undefined at \( x = 0 \).
Function \( h \) does not satisfy all conditions of Rolle's theorem.
d)
\( k(x) = |\sin(x)| \), for \( x \) in \( [0 , 2\pi] \)
The graph of function \( k \) is shown below and it shows that function \( k \) is not differentiable at \( x = \pi \).
Function \( k \) does not satisfy all conditions of Rolle's theorem.

Graph of k(x) = |sin(x)| , for x in [0 , 2π] — Figure 4. Graph of \( k(x) = |\sin(x)| \), for \( x \) in \( [0 , 2\pi ] \)

Question 2

Check that function \( f(x) = x^2 - 4x + 3 \) on the interval [1 , 3] satisfies all conditions of Rolle's theorem and then find all values of \( x = c \) such that \( f '(c) = 0 \).

solution to question 2

\( f \) is a polynomial function and is therefore continuous on the interval [1 , 3] and differentiable on the interval (1 , 3). Also \( f(1) = f(3) = 0 \) and therefore function \( f \) satisfies all three conditions of Rolle's theorem and there is at least one value of \( x = c \) such that \( f '(c) = 0 \).
\( f '(x) = 2x - 4 \)
\( f '(c) = 2c - 4 = 0 \)
Solve for \( c \) to obtain
\( c = 2 \).
The graph below shows \( f \), \( f' \) and the tangent at \( x = c = 2 \) is horizontal; and \( f'(2) = 0 \)

use of Rolle's theorem question 2, f(x) = x^2 - 4x + 3 — Figure 5. Rolle's theorem , question 2, \( f(x) = x^2 - 4x + 3 \)

Question 3

Check that function \( g(x) = \cos(x) \) on the interval \( [- \pi /2 , 3\pi/2] \) satisfies all conditions of Rolle's theorem and then find all values \( x = c \) such that \( g '(c) = 0 \).

solution to question 3

Function \( g \) is a cosine function and is therefore continuous on the interval \( [- \pi/2 , 3\pi/2] \) and differentiable on the interval \( (- \pi /2 , 3\pi /2) \) . Also \[ g(- \pi /2) = g( 3\pi /2) = 0 \] and therefore function \( g \) satisfies all three conditions of Rolle's theorem and there is at least one value of \( x = c \) such that \( f '(c) = 0 \).
\( g '(x) = - \sin(x) \)
\( g '(c) = - \sin(c) = 0 \)
Solve for \( c \) to obtain
\( c = n \pi \) , \( n = 0,\pm 1 , \pm 2 , ... \)
Solutions on the interval \( [- \pi/2 , 3\pi /2] \) are
\( c_1 = 0 \) and \( c_2 = \pi \);
The graph below shows \( g \), \( g' \) and the tangents at \( x = c_1 = 0 \) and \( x = c_2 = \pi \) are horizontal; and \( g'(0) = 0 \) and \( g'(\pi) = 0 \)

use of Rolle's theorem question 3 , g(x) = cos(x) — Figure 5. Rolle's theorem , question 3, \( g(x) = \cos(x) \)